Question

Calculate the concentrations of all species present (Na+, C6H5CO?2, C6H5CO2H, H3O+ and OH?) in 0.050 M sodium benzoate.

Express your answers using two significant figures. Enter your answers numerically separated by commas.

this is my table and some solutions pertaining to this problem

C6H5COO^-(aq)+H2O(l)<------->C6H5COOH(aq)+OH^-(aq)

I(M) .050 0 0

C -x +x +x

E .050-x x x

Kw/Ka= 10^-14/6.5*10^-5=1.538*10^-10 (not sure how we got the fraction if this can be explained that be great)

after solving we get= 2.77*10^-6M

can i get an explanation on how we find the M from H3O+ i did not even get this in my formula so im confused

and also OH- too.

Answer 1

0.050 M sodium benzoate

sodium benzoate is the salt of weak acid and stong base . so Na+ concentration = 0.050 M

[Na+] = 0.050 M

C6H5COO- + H2O ----------------------> C6H5COOH + OH-

0.050 0 0 --------------------> initial

-x + x + x -----------------> change

0.050-x x x ----------------------> equilibrium

Kb = [C6H5COOH][OH-]/[C6H5COO-]

Kb = x^2 / 0.050 -x

Kw / Ka =  x^2 / 0.050 -x

1.0 x 10^-14 / 6.5 x 10^-5 =  x^2 / 0.050 -x

1.538 x 10^-10 = x^2 / 0.050 -x

x^2 + 1.538 x 10^-10 x - 7.69 x 10^-12 = 0

by solving this equation

x = 2.77 x 10^-6

x = [C6H5COOH] = [OH-] = 2.8 x 10^-6 M

[C6H5COO-] = 0.050 -x = 0.050M

[H3O+] = Kw / [OH-]

= 1.0x 10^-14 / 2.77 x 10^-6

= 3.6 x 10^-9 M